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Alla [95]
3 years ago
14

A circle is translated 4 units to the right and then reflected over the x axis. What statement will always be true of the circle

at the new location?
Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

The statements are not given, so i will answer it in the most general way possible:

First, let's recall two things:

Horizontal translation.

For a number A positive, we can move the graph of f(x) by A units to the right by the transformation g(x) = f(x - A)

Reflection over x

For a point (x, y), a reflection over the x-axis transforms this point into (x, - y).

Now let's analyze the problem:

A generic circle, of radius R and centered in the point (a, b) can be written as:

(x - a)^2 + (y - b)^2 = R^2.

Now let's translate this 4 units to the right:

(x - 4 - a)^2 + (y - b)^2 = R^2.

or in standard circle notation:

(x - (4 + a))^2 + (y - b)^2 = R^2.

Now the circle is centered in the point (a + 4, b).

Now let's do a reflection over the x-axis.

This means that the sign of y changes, so now we have:

(x - (4 + a))^2 + (-y - b)^2 = R^2.

But because the term with y is squared, this is the same than:

(x - (4 + a))^2 + (y + b)^2 = R^2.

Or, in standard circle notation:

(x - (4 + a))^2 + (y - (-b))^2 = R^2.

Then the circle is now centered in point (a + 4, -b).

So now you know where the circle is located, and also you can see that the radius of the circle never changed.

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1<br> 1<br> 8 x 1,000) + (3 x 100) + (2 x 1) + (4 x 10) + (7 X 1000
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Step-by-step explanation:

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Gwen has $20, $10, and $5 bills in her purse worth a total of $220. She has 15 bills in all. There are 3 more $20 bills than the
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Answer:

x  =  8   ( 20$ bills)

y  = 5    ( 10 $ bills)

z = 2     ( 5  $  bills)

Step-by-step explanation:

Let call x, y, and z the number of bill of 20, 10, and 5 $ respectively

then according to problem statement, we can write

20*x + 10*y + 5*z = 220         (1)

We also know the total number of bills (15), then

x + y + z = 15     (2)

And that quantity of 20 $ bill is equal to

x = 3 + y     (3)

Now we got a three equation system we have to solve for x, y, and z for which we can use any valid procedure.

As    x = 3 + y    by substitution in equation (2)   and (1)

( 3 + y ) + y + z  = 15       ⇒   3 + 2*y + z = 15  ⇒  2*y + z = 12

20* ( 3 + y ) + 10*y + 5*z  = 220  ⇒ 60 + 20*y + 10*y + 5*z = 220

30*y + 5*z  = 160      (a)

Now we have only 2 equations

2*y + z = 12   ⇒    z = 12 - 2*y

30*y + 5*z  = 160     30*y  + 5* ( 12 - 2*y) = 160

30*y  + 60 - 10*y = 160

20*y = 100

y = 100/20       y = 5      Then by substitution in (a)

30*y + 5*z = 160

30*5  + 5*z = 160

150 + 5*z  = 160    ⇒     5*z = 10     z = 10/5      z = 2

And x

x + y + z = 15

x + 5 + 2 = 15

x = 8

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Answer:

c

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